Prove two formulas with partial derivatives

Show that

$x \left( \frac{\partial z}{\partial x} \right) + y \left( \frac{\partial z}{\partial y} \right) = 2z$

when

We compute,
$\begin{align*} x \left( \frac{\partial z}{\partial x} \right) + y \left( \frac{\partial z}{\partial y} \right) &= x \left( \frac{\partial}{\partial x} (x-2y)^2 \right) + y \left( \frac{\partial}{\partial y} (x-2y)^2 \right) \\ &= x (2 (x-2y)) + y (2(-2)(x-2y)) \\ &= 2x^2 - 4xy - 4xy + 8y^2 \\ &= 2 (x^2 - 4xy + 4y) \\ &= 2 (x - 2y)^2 \\ &= 2z. \end{align*}$
We compute,
$\begin{align*} x \left( \frac{\partial z}{\partial x} \right) + y \left( \frac{\partial z}{\partial y} \right) &= x \left( \frac{\partial}{\partial x} (x^4 + y^4)^{\frac{1}{2}} \right) + y \left( \frac{\partial}{\partial y} (x^4+y^4)^{\frac{1}{2}} \right) \\ &= x (4x^3)(1/2)(x^4+y^4)^{-\frac{1}{2}} + y (4y^3) (1/2) (x^4+y^4)^{-\frac{1}{2}} \\ &= \frac{2x^4 + 2y^4}{\sqrt{x^4+y^4}} \\ &= 2 (x^4+y^4)^{\frac{1}{2}} \\ &= 2z. \end{align*}$

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